Question

The sum of the first n terms of the series ${{1}^{2}}+{{2.2}^{2}}+{{3}^{2}}+{{2.4}^{2}}+{{5}^{2}}+{{2.6}^{2}}+\cdots $ is $\dfrac{n{{\left( n+1 \right)}^{2}}}{2}$, when n is even. When n is odd sum is:

Answer 1

Hint: According to the given question, we need to find the sum of the series when the number of terms is odd and when the number of terms are even then we are already provided the general term of the sum. Also, we need to keenly observe the given series and then proceed to find the sum.

Complete step-by-step solution:
In the question, we are given series as ${{1}^{2}}+{{2.2}^{2}}+{{3}^{2}}+{{2.4}^{2}}+{{5}^{2}}+{{2.6}^{2}}+\cdots $and we also know that the sum of the series is $\dfrac{n{{\left( n+1 \right)}^{2}}}{2}$where n is even this is provided.
Now, we need to find the sum of the same series but when n is odd. So, for that we need to let $n=2m+1$ . Now, the required sum would be \[\begin{align}
& {{1}^{2}}+{{2.2}^{2}}+{{3}^{2}}+{{2.4}^{2}}+{{5}^{2}}+{{2.6}^{2}}+\cdots +2{{\left( 2m \right)}^{2}}+{{\left( 2m+1 \right)}^{2}} \\
 & \Rightarrow \sum{{{\left( 2m+1 \right)}^{2}}+4\left( {{1}^{2}}+{{2}^{2}}+\cdots +{{m}^{2}} \right)} \\
\end{align}\]
Now, from this if we add odd place terms and even place terms then we get,
$\begin{align}
  & =\dfrac{\left( 2m+1 \right)\left( 2m+2 \right)\left( 4m+2+1 \right)}{6}+\dfrac{4m\left( m+1 \right)\left( 2m+1 \right)}{6} \\
 & \Rightarrow \dfrac{\left( 2m+1 \right)\left( m+1 \right)}{6}\left[ 2\left( 4m+3 \right)+4m \right] \\
 & \Rightarrow \dfrac{\left( 2m+1 \right)\left( 2m+2 \right)\left( 6m+3 \right)}{6} \\
 & \Rightarrow \dfrac{{{\left( 2m+1 \right)}^{2}}\left( 2m+2 \right)}{2} \\
\end{align}$
Now, we have almost attained the answer, further simplifying this would give us the final answer which is $\dfrac{{{n}^{2}}\left( n+1 \right)}{2}$ .
Therefore, the sum of given series when the number of terms is odd is given by $\dfrac{{{n}^{2}}\left( n+1 \right)}{2}$.

Note: In such a question we need to observe the given series carefully first. Since, most of the times more than one kind of series is mixed in alternate positions and hence we need to just identify that and then proceed in the same way as earlier but for all the different series in order to get the answer of one long series.